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Sub-problem 4: Both decide capacity and PC decides timing while TSO has

C.4.1 PC’s investment problem

When the PC has timing and sizing flexibility while the TSO has to invest at time zero, the investment problem of the PC is equal to:

sup

As in Sub-problem 3, the solution to this problem is equal to the one derived in Section 5.1.1. The optimal decision is to expand generation capacity with KP equal to:

KP(✓P) = max

where ˆ✓P is given by the solution to the following implicit equation:

✓ˆP = 1

1 1(⇢ ↵)

1 ⌘(2K0+ KP(✓P)). (C.55)

In the case where KP is equal to zero, generation capacity will never be added, i.e., ✓P(0)=1.

The optimal value of the endogenous constant A1 is equal to:

A1 = KP(✓P)(1 ⌘(2K0 + KP(✓P)))

⇢ ↵

1

1

✓ˆP⇤1 1. (C.56)

C.4.2 TSO’s investment problem

The TSO’s investment problem now only includes the flexibility to decide on capacity. At time zero the investment problem is equal to:

maxKT E

 Z 1

s=0

e ⇢sts(✓s; K0)ds (K0+ KT) +

Z 1

s=⌧P

e ⇢s[ts(✓s; K0+ KP) ts(✓s; K0)]ds e ⇢⌧P KT0 . (C.57)

The only way the TSO can influence the PC’s investment decision in this case is by investing in a capacity which is smaller than the PC’s optimal capacity. Making it invest in a smaller capacity than it finds optimal, will not only a↵ect the size of the PC’s investment, but also force it to invest earlier as its investment threshold is decreasing in capacity. We assume that the TSO has the power to a↵ect the PC’s investment decision. Then we need to substitute the expression for ✓P(KT) into the total surplus expression, before we maximise it with respect to KT. We get the same expression for KT, Equation (59), as in Section 5.1.2. KT is equal to:

KT(✓0) = max ˆKT(✓0), 0 , (C.58)

where ˆKT(✓0) is given by the following implicit equation:

+

 [2⌘ ˆKT(⌘K0( 2) + 1) + (2⌘K0 1)(2⌘K0( 2) + 1])

2( 1)(⌘ ˆKT + 2⌘K0 1)2 +

1 2 ⇤

( 1)✓0(⌘ ˆKT + 2⌘K0 1)

(↵ ⇢) = 0. (C.59)

C.4.3 Optimal investment strategies

To find the optimal investment strategies after taking capacity constraints in to account, we need to compare KT and KP to determine which one of them will be dominating. If KT > KP, the PC will expand its capacity with KP when ✓t reaches ✓P(KP), and the TSO will have to accept the PC’s optimal investment strategy and that it has no power to a↵ect it. However, if KT < KP the PC will have to accept the optimal capacity of the TSO, and choose it’s optimal investment time based on KT, ✓P(KT). This optimal investment time will be the same as the TSO took into account when it found its optimal KT. The optimal investment strategies when KT is the dominating capacity is summarised in Table C.3 while

the optimal investment strategies when KP is the dominating capacity is summarised in

Table C.3: Overview of optimal investment strategies for Sub-problem 4 if KT is the domi-nating capacity

Given the dominating capacity, min(KT, KP), and the resulting investment threshold for the PC, the social welfare at time zero will be equal to:

VT SO(✓0, ✓P, min(KT, KP)) =h1

Optimal investment strategies Sub-problem 4

Table C.4: Overview of optimal investment strategies for Sub-problem 4 if KP is the domi-nating capacity

and the value for the PC at time zero will be equal to:

VP C(✓0, ✓P, min(KT, KP)) = h In the special case where KT or KP is equal to zero, the PC will never expand its capacity above K0, the present value for the TSO and the PC, given that the TSO invests in K0 at time ✓T, is equal to:

C.5 Sub-problem 5: Both have sizing flexibility, while the TSO decides timing

C.5.1 PC’s investment problem

When the PC only has sizing flexibility, and has to follow the TSO’s timing decision, the investment problem of the PC at time zero is equal to:

maxKP E Z 1

s=⌧T

e ⇢s⇡(✓s, K0+ KP)ds e ⇢⌧P KP0 . (C.64)

The solution to this problem is equal to Equation (13) but with ✓T instead of ✓t: KP(✓T) = max

When the PC has to invest at the same time as the TSO, the TSO’s investment problem at time zero is equal to:

If both invest at ✓t, the total expected surplus at time t is given by:

T S(✓t, KT) =

1

2⌘(K0+ KT)2+ (1 ⌘(K0+ KT))(K0+ KT) ✓t

⇢ ↵ (K0+ KT) KT. (C.67)

The solution to the inner maximisation problem is equal to the one in Sub-problem 1, see Appendix C.1 but with but with ✓t instead of ✓0:

After having solved for KT, the TSO’s investment problem reduces to:

sup

T E

 Z 1

s=⌧T

e ⇢sts(✓s; K0 + KT)ds (K0+ KT) KT0 . (C.69)

We proceed by following a dynamic programming approach to solve the optimal stopping problem and find ✓T. The value for the TSO at time t is equal to:

F (✓t, KT(✓t)) = 8<

:

B1t1 if ✓t ✓T,

T S(✓t, KT(✓t)) if ✓T  ✓t. (C.70)

The value in the continuation region is equal to the value of the option to invest in the transmission line, while the value in the stopping region is equal to the value of the total surplus given that they both have invested. The value in the continuation region is derived by finding the solution to the ordinary di↵erential equation (ODE) that stews from the Bellman equation:

⇢F dt =E[dF ]. (C.71)

Standard calculations similar to those in Section 5.1.1 and 5.1.2 are performed, which lead to the value function stated in Equation (C.70).

The second branch is equal to:

T S(✓t, KT(✓t)) =

1

2⌘(K0+ KT(✓t))2 + (1 ⌘(K0+ KT(✓t)))(K0+ KT(✓t)) ✓t

⇢ ↵

(K0+ KT(✓t)) KT(✓t). (C.72)

To determine the investment threshold ✓T and the value of the endogenous constant B1, we employ the value-matching and smooth-pasting conditions. First, the value-matching condition is given by:

B1T 1 = T S(✓T, KT(✓T)). (C.73)

When deriving the smooth-pasting condition, we have to take into account that KT depends on ✓T. We get:

B1 1T 1 1 = @T S

@✓t t=✓T

+@T S

@KT

@KT

@✓t t=✓T

. (C.74)

However after maximising the total surplus (TS) with respect to KT, we have by the envelope theorem that @K@T S

T = 0. Therefore the smooth-pasting condition reduces to:

B1 1T 1 1 = @T S

@✓t t=✓T

. (C.75)

The smooth-pasting condition gives:

B1 =

(K0+ KT(✓T))h

1 1

2⌘(K0+ KT(✓T))i ✓⇤1T 1

1(⇢ ↵). (C.76)

Substituting the expression for B1 into the value-matching condition and solving for ✓T, we get that ✓T is given by the solution to the following implicit equation:

T = 1

1 1(⇢ ↵) (K0 + KT(✓T)) + KT(✓T)

(K0+ KT(✓T))[1 12⌘(K0+ KT(✓T))] (C.77) C.5.3 Optimal investment strategies

The final optimal investment strategy depends on which agent that has the lowest optimal capacity. If KP is larger than KT, the TSO will decide the timing and the size of both investments. However, if the opposite holds, that KT is larger than KP, the PC decides the size of both investments, while the TSO decides the timing. The optimal investment strategies when KT is the dominating capacity is summarised in Table C.5 while the optimal investment strategies when KP is the dominating capacity is summarised in Table C.6.

Optimal investment strategies Sub-problem 5 KT < KP

KP = KTP = ✓T

TSO’s optimal strategy:

KT = max

1

1 ( + )(⇢ ↵)

T K0, 0

T = 111(⇢ ↵)(K (K0+KT(✓T))+ KT(✓T)

0+KT(✓T))[1 12⌘(K0+KT(✓T))]

PC’s optimal decision:

KP = KT

P = ✓T

Table C.5: Overview of optimal investment strategies for Sub-problem 5 if KP is the domi-nating capacity

Optimal investment strategies Sub-problem 5

Table C.6: Overview of optimal investment strategies for Sub-problem 5 if KT is the domi-nating capacity

The resulting social welfare for the TSO will be:

VT SO(✓T, ✓T, min(KT, KP)) =

And the value of the PC:

VP C(✓T, ✓T, min(KT, KP)) =